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Discussion Forum



1)

Let [ε0] denote the dimensional formula of the permittivity of vacuum. If M= mass, L= length , T= time and A= electric current then,

  


A) $[\epsilon_{0}]=[M^{-1}L^{-3}T^{2}A^{}]$

B) $[\epsilon_{0}]=[M^{-1}L^{-3}T^{4}A^{2}]$

C) $[\epsilon_{0}]=[M^{-2}L^{2}T^{-1}A^{-2}]$

D) $[\epsilon_{0}]=[M^{-1}L^{2}T^{-1}A^{2}]$

Answer:

Option B

Explanation:

from COULOMB'S law  

                           $F=\frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{R^{2}}$

$\epsilon_{0}=\frac{q_{1}q_{2}}{4\pi F R^{2}}$

substituting the units we have ,

 $\epsilon_{0}=\frac{C^{2}}{N.m^{2}}=\frac{[AT]^{2}}{[ML^{}T^{-2}][L^{2}]}=$

                                         $[M^{-1}L^{-3}T^{4}A^{2}]$